fourier analytic proofs - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:45:59Z http://mathoverflow.net/feeds/question/120485 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120485/fourier-analytic-proofs fourier analytic proofs Koushik 2013-02-01T03:46:15Z 2013-02-02T16:40:21Z <p>While searching through Mathoverflow, I found out a fourier analytic proof of the Isoperimetric Inequality.Also, by google search I found a fourier analytic proof of Quadratic Reciprocity theorem.I know of the fourier analytic approach used by combinatorialists like Ben Green. But what are the other fourier analytic proofs of some of the well known classical theorems other than what I have mentioned above specially those which admit a starkly different proofs.</p> http://mathoverflow.net/questions/120485/fourier-analytic-proofs/120498#120498 Answer by jbc for fourier analytic proofs jbc 2013-02-01T06:57:32Z 2013-02-01T06:57:32Z <p>There are two very extensive monographs on this subject, more precisely, its relationship to convex geometry: by Groemer (Geometric applications of Fourier series and spherical harmonics) and Koldobsky (Fourier analysis in convex geometry).</p> http://mathoverflow.net/questions/120485/fourier-analytic-proofs/120530#120530 Answer by Bazin for fourier analytic proofs Bazin 2013-02-01T16:53:51Z 2013-02-01T16:53:51Z <p>Let me speak about the "Triumph of Fourier" according to the words of Laurent Schwartz in his autobiography. The Fourier transformation is a handy tool to characterize regularity of functions. </p> <p>Let $u$ be a distribution on some open subset $\Omega$ of $\mathbb R^n$. A consequence of the Paley-Wiener theorem is that a point $x_0$ in $\Omega$ is not in the singular support of $u$ whenever there exists a neighbordhood $U$ of $x_0$ such that $$\forall \chi\in C_c^\infty(U), \forall N\in \mathbb N,\quad\vert\widehat{\chi u}(\xi)\vert\vert \xi\vert^N\in L^{\infty}(\mathbb R^n).$$</p> <p>That notion can be refined to define the ($C^\infty$) wave-front-set, as a subset of the cotangent bundle (minus the 0 section): a point $(x_0,\xi_0)\in \Omega \times\mathbb S^{n-1}$ does not belong to the wave-front-set of $u$ whenever there exists a neighbordhood $U$ of $x_0$, a neighborhood $V$ of $\xi_0$ on the sphere such that $$\forall \chi\in C_c^\infty(U), \forall N\in \mathbb N,\quad\vert\widehat{\chi u}(t\xi)\vert t^N\in L^{\infty}((1,+\infty)_t\times V).$$</p> <p>The wave-front-set (WF) can be used to detect the various directions of singularities: for instance $$WF(\delta_{0})=\text{{0}}\times (\mathbb R^n\backslash\text{{0}})$$ but with $H=\mathbf 1_{(0,+\infty)}$ the Heaviside function, $H(x_1)$ in $\mathbb R^n$ is also singular at 0 but the structure of the singularity is quite different and indeed with $\Sigma=${$x\in \mathbb R^n, x_1=0$} $$WF(H(x_1))=\Sigma\times\text{{0\not=\xi\in \mathbb R^n, \xi_2=\dots=\xi_n=0}}$$ which is the conormal bundle to $\Sigma$. The first projection of the wave-front-set is the singular support.</p> <p>That definition can be extended to Sobolev regularity (spaces based on $L^2$), analytic regularity (more generally Gevrey) and is the only way to express the propagation of singularities for linear waves.</p> http://mathoverflow.net/questions/120485/fourier-analytic-proofs/120531#120531 Answer by Nik Weaver for fourier analytic proofs Nik Weaver 2013-02-01T18:19:56Z 2013-02-01T18:19:56Z <p>Hermann Weyl's delightful proof that for irrational $\alpha$ the sequence of values $k\alpha$ mod $1$, $k \in {\bf N}$, is uniformly distributed in $[0,1]$ deserves a mention. It's so simple I can summarize it here. First we check that for any nonzero $n \in {\bf Z}$ we have $$\frac{1 + e^{2\pi i n\alpha} + \cdots + e^{2\pi in(k-1)\alpha}}{k} \to 0$$ as $k \to \infty$. This is just a simple computation since the numerator is a geometric series. For $n = 0$ the displayed fraction reduces to $\frac{k}{k} = 1$. Since $\int_0^1 e^{2\pi i nx} dx = 1$ or $0$ depending on whether $n = 0$ or $n \neq 0$, it follows that $$\frac{1}{k}\sum_{j=0}^{k-1} e^{2\pi i nj\alpha} \to \int_0^1 e^{2\pi inx} dx$$ for all $n \in {\bf Z}$. Setting $x_j = j\alpha$ mod $1$ and taking linear combinations then yields $$\frac{1}{k}\sum_{j=0}^{k=1} f(x_j) \to \int_0^1 f(x) dx$$ for any trigonometric polynomial $f$, and by straightforward approximation arguments we get the same conclusion, first for any continuous function $f$ on $[0,1]$ and then for $f = \chi_{[a,b]}$. But with this $f$ the left side becomes the fraction of values $j\alpha$ mod $1$ for $0 \leq j \leq k-1$ which lie in $[a,b]$ and the right side becomes $b-a$, so this is just the statement of uniform distribution.</p> http://mathoverflow.net/questions/120485/fourier-analytic-proofs/120601#120601 Answer by Noam D. Elkies for fourier analytic proofs Noam D. Elkies 2013-02-02T16:40:21Z 2013-02-02T16:40:21Z <p>The sign of the quadratic Gauss sum $\tau$ can be obtained from the spectrum of the discrete Fourier transform $\Phi$: the trace of $\Phi$ gives $\tau$, and $\det\Phi$ distinguishes $\tau$ from $-\tau$.</p> <p>Recall that for an odd prime $p$ the <em>quadratic Gauss sum</em> can be defined by $$\tau = \sum_{n=0}^{p-1} \zeta^{n^2}$$ where $\zeta = e^{2\pi i / p}$. It is elementary that $|\tau|^2 = p$ and that $\tau$ is real or pure imaginary according as $p \equiv 1 \bmod 4$ or $p \equiv -1 \bmod 4$. In fact $\tau$ is always $+\sqrt p$ in the former case, and $+i\sqrt p$ in the latter, but this is notoriously tricky to prove.</p> <p>One trick is to recognize $\tau$ as the trace of the discrete Fourier transform on ${\bf C}^p$, which has matrix $$\Phi = (\zeta^{mn})_{m,n=0}^{p-1}.$$ Now $\Phi^2$ is the matrix whose $(m,n)$ entry is $p$ if $m+n \equiv 0 \bmod p$ and $0$ otherwise (this is tantamount to discrete Fourier inversion). This matrix has eiganvalues $+1$ and $-1$ with multiplicity $(p+1)/2$ and $(p-1)/2$ respectively. Hence $\Phi$ has eigenvalues $i^k \sqrt p$ ($k=0,1,2,3$) with multiplicities $m_k$ satisfying $m_0 + m_2 = (p+1)/2$ and $m_1 + m_3 = (p-1)/2$, and then $\tau = \sum_{k=0}^3 m_k i^k \sqrt p$. Since we already know $\tau$ up to sign there are only two possibilities: if $p \equiv 1 \bmod 4$ then $m_0$ or $m_2$ is $(p+3)/4$ and the other three $m_k$ are $(p-1)/4$, while if $p \equiv -1 \bmod 4$ then $m_1$ or $m_3$ is $(p-3)/4$ and the other three $m_k$ are $(p+1)/4$. We are to show that the odd man out is always $m_0$ in the former case and $m_3$ in the latter.</p> <p>In each case we can decide the correct choice by computing the sign (a.k.a. argument) of $\det \Phi = p^{p/2} \prod_{k=0}^3 i^{k m_k}$. We can do this because $\Phi$ is a Vandermonde matrix, whence $\det\Phi$ has the product expansion $\prod_{0 \leq m &lt; n &lt; p} (\zeta^n - \zeta^m)$. Each factor $\zeta^n - \zeta^m$ is a positive real multiple of $\exp((m+n+\frac12)\pi i)$. It soon follows that $\det\Phi = i^{(1-p)/2} p^{p/2}$ (we already knew $\left|\det\Phi\right|$ because each eigenvalue has absolute value $\sqrt{p}$), and conclude as desired that $\tau = \sqrt{p}$ when $p \equiv 1 \bmod 4$ while $\tau = i\sqrt{p}$ when $p \equiv -1 \bmod 4$.</p> <p>[This looks like a known but not very well-known argument that is easier to rediscover than to find in the literature. What is the original source?]</p>